Smooth deformation of a smooth curve may result in a singularity, example given by Graham.

Extra

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%The last two examples, are the minimal non trivial such examples.\\
The above example could be applied for any scheme in the sense of \ref{Scheme}, as well as $D((f))\neq \emptyset$, otherwise $\rho_0,\rho_1$ might be the same. \tcb{another case is when $A=\KK$ is a field}.
%\begin{Qs}
%Would the inverse of the last mark be true? The answer is NOT the below proposition.
%\end{Qs}

Discrete topology. Therefore, we might use it explicitly
natural example
other sheaf on ringed space

Bloch Conjuncture
Hodge Conjuncture
be a category with all products (particularly it has a terminal object $T$)
Discrete topology.z

\begin{Rem}
\tcb{In the definition of the sieve, this condition is to say that we require the restriction of the shieve not to depend on the path of restriction! But about gluing!}
\end{Rem}
===== Is it that we to consider the pretopology then we sievefy it!?

\begin{Rem}
\tcb{Show that it is a pre-topology!} Actually it is not because $id_X$ is not a sieve in general.
\end{Rem}


\begin{Qs}
\tcr{Did Grothendieck pre-topology was defined based on the notion of covering only because of the goal of defining sheaves? Or is it the most natural approach to provide a categorical definition of confiontional topological spaces?}
\end{Qs}
First, Answer is No

\begin{Ex}
Let $\goC$ be the preordered category of open subsets of the topological space $(X,\tau_X)$. Then, we may define a Grothendieck covering that coincide with the topological open covering, i.e. we say that $\{f_{\alpha}:U_{\alpha}\rightarrow U:\alpha\in A\}$ is a covering for $U\in \goC$ iff $U=\displaystyle\cup_{\alpha\in A}U_{\alpha}$. One sees readily that it defines a Grothendieck pre-topology.
\end{Ex}

Different diagrams, same deduced arrows\\
maximal equality!!

\begin{Ex}[Sheaf, illustrating a case when $\rho_0\neq \rho_1$]
Let $A=\KK[x]$ , for some algebraically closed field $\KK$, and let $((X,\tau),\OC)=(\Spec A,\tau_{\Spec A},\OC_{\Spec A})$. Then for some covering of $U\subseteq X$ the above defined $\rho_0,\rho_1$ are two distinct morphisms.
\end{Ex}
\begin{proof}
One can readily show $(\tau,\subseteq)$ defines a Grothendieck pre-topology, then $\emptyset \neq D((x))\subsetneqq X$. Let $U_1=D((x)),U_2=X$, then we have $\{f_1=i:D((x))\rightarrow X, f_2=id_X:X\rightarrow X \}$ a covering for $X$. 
$\OC$ is a contravariant functor, then there is a diagram $A=\OC(X)\stackrel{id_A}{\leftarrow} A=\OC(X) \stackrel{i^{\#}}{\rightarrow} A_x=i_{\ast} \OC{d((x))}$. Hence, $\exists! \rho:A\rightarrow A_x\times A$ that makes the product diagram commutative, and it is given by $\rho(x)=(x,x)$.\\
We notice that $D((x))\times_x X=D((x)), D((x))\times_x D((x))=D((x))$, and $X\times_x X=X$ (accompanied with inclusions as its projection), then the following four diagram commute:\\
$\xymatrix{D((x))\ar@[-->]@[blue][r]^{id_{D((x))}}\ar@[-->]@[blue][d]_{id_{D((x))}}&D((x))\ar[d]^i\\
D((x)) \ar[r]_{i} & X}$
$\xymatrix{D((x))\ar@[-->]@[blue][r]^{id_{D((x))}}\ar@[-->]@[blue][d]_{i}&D((x))\ar[d]^i\\
X \ar[r]_{id_X} & X}$
$\xymatrix{D((x))\ar@[-->]@[blue][r]^{i}\ar@[-->]@[blue][d]_{id_{D((x))}}&X\ar[d]^{id_X}\\
D((x)) \ar[r]_{i} & X}$
$\xymatrix{X\ar@[-->]@[blue][r]^{id_X}\ar@[-->]@[blue][d]_{id_X}&X\ar[d]^{id_X}\\
X \ar[r]_{id_X} & X}$\\
$\OC$ is a contravariant functor, then the following diagram commute:\\
$\xymatrix{\OC(U_1\times_X U_1)=A_x&&A_x\ar@[-->]@[blue][ll]_{id_{A_x}}\\
A_x\ar@[-->]@[blue][u]^{id_{A_x}}  && A\ar[u]_i\ar[ll]^{i}}$\  
$\xymatrix{\OC(U_2\times_X U_1)=A_x&&A_x \ar@[-->]@[blue][ll]_{id_{A_x}}\\
A \ar@[-->]@[blue][u]^{i} && A\ar[u]_i \ar[ll]^{id_A}} $\\  
$\xymatrix{\OC(U_1\times_X U_2)=A_x&&A \ar@[-->]@[blue][ll]_{i}\\
A_x\ar@[-->]@[blue][u]^{id_{A_x}}  && A\ar[u]_{id_A}\ar[ll]^{i}}$
$\xymatrix{\OC(U_2\times_X U_2)=A&&A \ar@[-->]@[blue][ll]_{i}\\
A \ar@[-->]@[blue][u]^{i} && A\ar[u]_{id_A}\ar[ll]^{id_A}}$\\
Then the first raw morphisms in the above diagrams induce the morphisms:
\[\begin{array}{lll}
A_x:&\rightarrow & A_x\times A_x=\OC(U_1\times_X U_1)\times \OC(U_2\times_X U_1)\\
y&\mapsto& (y,y)
\end{array}\]\[
\begin{array}{lll}
A:&\rightarrow & A_x\times A=\OC(U_1\times_X U_2)\times \OC(U_2\times_X U_2)\\
z&\mapsto& (z,z)
\end{array}\]
Which in tern induce the morphism:\\
\[\begin{array}{lcll}
\rho_0:&\OC(U_1)\times \OC(U_2):&\rightarrow & \OC(U_1\times_X U_1)\times \OC(U_2\times_X U_1)\times \OC(U_1\times_X U_2)\times \OC(U_2\times_X U_2)\\
&(y,z)&\mapsto& (y,y,z,z)
\end{array}\]
On the other hand, the first raw morphisms in the above diagrams induce the morphisms:
\[\begin{array}{lll}
A_x:&\rightarrow & A_x\times A_x=\OC(U_1\times_X U_1)\times \OC(U_1\times_X U_2)\\
y&\mapsto& (y,y)
\end{array}\]\[
\begin{array}{lll}
A:&\rightarrow & A_x\times A=\OC(U_2\times_X U_1)\times \OC(U_2\times_X U_2)\\
z&\mapsto& (z,z)
\end{array}\]
Which in tern induce the morphism:\\
\[\begin{array}{lcll}
\rho_1&\OC(U_1)\times \OC(U_2):&\rightarrow & \OC(U_1\times_X U_1)\times \OC(U_2\times_X U_1)\times \OC(U_1\times_X U_2)\times \OC(U_2\times_X U_2)\\
&(y,z)&\mapsto& (y,z,y,z)
\end{array}\]
Then one can readily see that $\rho$ is the equaliser of the distinct $\rho_0,\rho_1$.
\end{proof}
